A polarizer-analyser set is a adjusted such that the intensity of light coming out of the analyser is just 12.5 % of the original intersity. Assuming that the polarizer – analyser set does not absorb any light the angle by which the analyser need to be rotated further to reduce the output intensity to be zero, is :

Options

A. `45^(@)`
B. `90^(@)`
C. `60^(@)`
D. `30^(@)`

Explanation

To solve the problem, we need to find the angle by which the analyser needs to be rotated further to reduce the output intensity of light to zero. We will use Malus's Law, which states that the intensity of light passing through a polarizer-analyzer system is given by:

\[ I = I_0 \cos^2(\theta) \]

where:
- \( I \) is the transmitted intensity,
- \( I_0 \) is the initial intensity of light,
- \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The intensity of light coming out of the analyser is \( 12.5\% \) of the original intensity \( I_0 \).
- This means:
\[ I = 0.125 I_0 \]

2. **Using Malus's Law:**
- According to Malus's Law:
\[ I = I_0 \cos^2(\theta) \]
- We can substitute for \( I \):
\[ 0.125 I_0 = I_0 \cos^2(\theta) \]

3. **Simplifying the Equation:**
- Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 0.125 = \cos^2(\theta) \]

4. **Finding \( \theta \):**
- Taking the square root of both sides:
\[ \cos(\theta) = \sqrt{0.125} = \frac{1}{2\sqrt{2}} \approx 0.3536 \]
- Now, we find \( \theta \):
\[ \theta = \cos^{-1}(0.3536) \approx 75^\circ \]

5. **Finding the Angle for Zero Intensity:**
- To reduce the intensity to zero, we need to rotate the analyser to \( 90^\circ \) relative to the polarizer.
- The angle \( \theta \) we found is \( 75^\circ \), so the additional angle \( \Delta \theta \) needed to rotate is:
\[ \Delta \theta = 90^\circ - 75^\circ = 15^\circ \]

### Final Answer:
The angle by which the analyser needs to be rotated further to reduce the output intensity to zero is **15 degrees**.